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Find the value of \(x = 1 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \ddots}}}}\)

 

Hint(s):What's the relationship between \(x\) and \(x+1\)?

 Dec 21, 2018
 #1
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Here is a computer summation of your continued fraction:
=1E100; listforeach(b,reverse(1,2,2,2.........etc ), c=b + 1/c);printc
c=1.4142135623730950488016887242097

The result is sqrt(2)

 Dec 21, 2018
edited by Guest  Dec 21, 2018
edited by Guest  Dec 21, 2018
 #2
avatar+100802 
+2

 

 

\(x = 1 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \ddots}}}}\\ x = -1+2 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \ddots}}}}\\ x +1= 2 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \ddots}}}}\\ x +1= 2 + \cfrac{1}{x+1}\\ \)

 

 

\(x +1= 2 + \cfrac{1}{x+1}\\ x-1= \cfrac{1}{x+1}\\ x^2-1=1\\ x^2=2\\ x=\pm\sqrt2 \)

 

I do not know why the answer cannot be -sqrt2  but if memory serves me correctly the only correct answer is +sqrt2

I'll continure thinking about that one.

 Dec 21, 2018
 #3
avatar+100516 
+2

It cannot be negative because the right side of the original equation is positive......thus...x is positive

 

 

cool cool  cool

CPhill  Dec 21, 2018
 #4
avatar+100802 
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Yes, I guess so.  Thanks Chris. 

Melody  Dec 22, 2018

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